Solve the differential equation- 1-x2 dy-dx-2xy- x 1-x2 1-2

The correct option is D y=√( ( 1 x^2 ))+c ( 1 x^2 ) #160;( 1 x^ 2 ) dy / dx +2xy=x( 1 x^ 2 ) ^ 1/2 ⇒ dy / dx + 2xy /( 1 x^ 2 ) #160; = x /√(( 1 ...

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Standard XII

Mathematics

Solving Linear Differential Equations of First Order

Solve the dif...

Question

Solve the differential equation: $(1 - x^{2}) \frac{d y}{d x} + 2 x y = x {(1 - x^{2})}^{1 / 2}$

y = \sqrt{(1 - x^{2})} + c (1 + x^{2})

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y = \sqrt{(1 + x^{2})} - c (1 + x^{2})

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y = \sqrt{(1 + x^{2})} + c (1 - x^{2})

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y = \sqrt{(1 - x^{2})} + c (1 - x^{2})

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Solution

The correct option is D $y = \sqrt{(1 - x^{2})} + c (1 - x^{2})$
$(1 - x^{2}) \frac{d y}{d x} + 2 x y = x {(1 - x^{2})}^{1 / 2} \Rightarrow \frac{d y}{d x} + \frac{2 x y}{(1 - x^{2})} = \frac{x}{\sqrt{(1 - x^{2})}}$ ..(1)
Here $P = \frac{2 x}{(1 - x^{2})} \Rightarrow \int P d x = \int \frac{2 x}{(1 - x^{2})} d x = - log (x^{2} - 1) = log \frac{1}{(x^{2} - 1)}$
$∴ I . F . = e^{log \frac{1}{(x^{2} - 1)}} = \frac{1}{(x^{2} - 1)}$
Multiplying (1) by $I . F .$ we get
$\frac{1}{(x^{2} - 1)} \frac{d y}{d x} - \frac{2 x y}{{(1 - x^{2})}^{2}} = \frac{x}{{(1 - x^{2})}^{3 / 2}}$
Integrating both sides we get
$\frac{y}{(x^{2} - 1)} = - \frac{1}{\sqrt{(1 - x^{2})}} + c \Rightarrow y = \sqrt{(1 - x^{2})} + c (1 - x^{2})$

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(b) x² + 1 = C (1 − y²)
(c) x³ − 1 = C (1 + y³)
(d) x³ + 1 = C (1 − y³)

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Q. STATEMENT -1 : Solution of

(1 + x \sqrt{x^{2} + y^{2}}) d x + y (- 1 + \sqrt{x^{2} + y^{2}}) d y = 0

x - \frac{y^{2}}{2} + \frac{1}{3} (x^{2} + y^{2})^{3 / 2} + C_{1} = 0, C_{1}

being arbitrary constant.
STATEMENT-2 : Solution of

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(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$

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Solve the differential equation: (1−x2)dydx+2xy=x(1−x2)1/2

Solve the differential equation: $(1 - x^{2}) \frac{d y}{d x} + 2 x y = x {(1 - x^{2})}^{1 / 2}$